题目链接
这道题会暴露很多孩纸的高数学的好不好。如何计算三维直线的最短距离和垂足的坐标，学习链接
之后的就是代入公式计算了。
1 #include <stdio.h> 2 #include <math.h> 3 struct Point { 4double x, y, z; 5Point(double x = 0, double y = 0, double z = 0):x(x), y(y), z(z) {} 6void read() { 7 scanf("%lf%lf%lf", &x, &y, &z); 8} 9void write() {10 printf("%.6lf%.6lf%.6lf", x, y, z);11}12Point operator + (const Point &b) {13 return Point(x + b.x, y + b.y, z + b.z);14}15Point operator - (const Point &b) {16 return Point(x - b.x, y - b.y, z - b.z);17}18Point operator * (double k) {19 return Point(x * k, y * k, z * k);20}21Point operator / (double k) {22 return Point(x / k, y / k, z / k);23}24Point cross(const Point &b) {25 return Point(y * b.z - z * b.y, z * b.x - x * b.z, x * b.y - y * b.x);26}27double len() {28 return sqrt(x * x + y * y + z * z);29}30double operator * (const Point &b) {31 return x * b.x + y * b.y + z * b.z;32}33 };34 typedef Point Vector;35 int main() {36int T;37scanf("%d", &T);38while (T--) {39 Point A, B, C, D;40 A.read();41 B.read();42 C.read();43 D.read();44 Vector AB= B - A, CD = D - C;45 Vector n = AB.cross(CD), AC = C - A, CA = A - C;46 double nlen = n.len(), dis = n * AC / nlen;47 double t1 = AC.cross(CD) * AB.cross(CD) / nlen / nlen;48 double t2 = CA.cross(AB) * CD.cross(AB) / nlen / nlen;49 Point E = A + (AB * t1);50 Point F = C + (CD * t2);51 if (dis < 0)52 dis = -dis;53 printf("%.6lf\n", dis);54 E.write();55 putchar(' ');56 F.write();57 putchar('\n');58}59return 0;60 }
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